A) \[0\]
B) \[-\frac{1}{\pi }\]
C) \[\frac{1}{\pi }\]
D) \[\frac{2}{\pi }\]
Correct Answer: D
Solution :
\[I=\int_{0}^{1}{|\sin 2\pi x|dx}\] \[=\int_{0}^{1/2}{(\sin 2\pi x)dx+\int_{1/2}^{1}{-(\sin 2\pi x)dx}}\] \[=-\frac{1}{2\pi }[\cos 2\pi x]_{0}^{1/2}+\frac{1}{2\pi }[\cos 2\pi x]_{1/2}^{1}=\frac{2}{\pi }\]
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